Problem: Region $R$ in the first quadrant is enclosed by the $y$ -axis, the line $y=4$, and the curve $y=x^2$. $y$ $x$ ${y=x^2}$ ${y=4}$ $ 0$ $ 2$ $ R$ What is the volume of the solid generated when $R$ is rotated about the $x$ -axis? Give an exact answer in terms of $\pi$.
Let's imagine the solid is made out of many thin slices. Each slice is a cylinder with a hole in the middle, much like a washer. $y$ $x$ ${y=x^2}$ ${y=4}$ $ 0$ $ 2$ Let the width of each slice be $dx$, let the radius of the washer, as a function of $x$, be $r_1(x)$, and let the radius of the hole, as a function of $x$, be $r_2(x)$. Then, the volume of each slice is $\pi[(r_1(x))^2-(r_2(x))^2]\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small width using a definite integral: $\int_a^b \pi [(r_1(x))^2-(r_2(x))^2]\,dx$ We call this the washer method. What we now need is to figure out the expressions for $r_1(x)$ and $r_2(x)$ and the interval of integration. $r_1(x)$ is equal to the distance from line $y=4$ to the $x$ -axis. So, $r_1(x)=4}$. $r_2(x)$ is equal to the distance between the curve $y=x^2$ and the $x$ -axis. So, ${r_2(x)=x^2}$. Now we can find an expression for the area of the washer's base: $\begin{aligned} &\phantom{=} \pi [(r_1(x)})^2-({r_2(x)})^2] \\\\ &= \pi [({4})^2-({x^2})^2] \\\\ &=\pi(16-x^4) \end{aligned}$ The leftmost endpoint of $R$ is at $x=0$ and the rightmost endpoint is at $x=2$. So the interval of integration is $[0,2]$. Now we can express the definite integral in its entirety! $\int_0^2 \left[\pi\left(16-x^4\right)\right]dx$ Let's evaluate the integral. $\int_0^2 \left[\pi\left(16-x^4\right)\right]dx=\dfrac{128\pi}{5}$ In conclusion, the volume of the solid is $\dfrac{128\pi}{5}$.